Diamond tradeoff: Difference between revisions

Line 49:

\end{bmatrix}

</math>

One property of an unchanged interval of a tuning is that it is [https://mathworld.wolfram.com/Eigenvector.html eigenvector] of the projection matrix<ref>For now, the best explanation of projection matrices seems to be on the [[fractional monzo]]s page.</ref> for the tuning where the eigenvalue <math>λ</math> is 1. If the projection matrix is <math>P</math>, by the definition of eigenvectors, that means <math>P⋅U = λ⋅U</math>, or <math>P⋅U = 1⋅U</math>, or simply <math>P⋅U = U</math>. In other words, the projection matrix maps the interval to itself; it is unchanged by the tuning. Because we know what <math>U</math> is, we could solve for <math>P</math> now. But we don't want <math>P</math>; we want the generators. Fortunately, <math>P</math> is defined in terms of our desired generators, <math>G</math>, and our mapping, <math>M</math>, like this: <math>P = GM</math>. Now, <math>G</math> is the [https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse pseudoinverse] of <math>M</math><ref>we can understand this fact in this context to mean that when you multiply <math>M</math> and <math>G</math> the other way around, <math>MG</math>, you get <math>I</math>, the identity matrix, and this is because <math>G</math> represents the generators in terms of the primes, just like any ordinary prime count vector ([[monzo]]), except often using fractional powers, so we map it with a mapping just like any ordinary interval, and if you look at the columns of <math>I</math> and think of them as generator count vectors, you can see that it represents our targets. In other words, <math>G</math> is the prime count vectors for which each one gets mapped to a different single one of the temperament's generators.</ref>; we often write the pseudoinverse of <math>M</math> as <math>M⁺</math>, so let's write <math>G</math> instead as <math>M⁺</math>. So if <math>P = M⁺⋅M</math>, then we can substitute that in for <math>P</math>, and our equation will now be <math>M⁺⋅M⋅U = U</math>. But we want to solve for our generators, so that's <math>M⁺</math>. So if we right multiply both sides by the inverse of <math>(M⋅U)</math>, we get <math>M⁺ = U(MU)^{-1}</math>; the <math>MU</math> on the left side cancels out. The rest is busywork.

We multiply <math>M⋅U</math>:

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</math>

Then find <math>T</math> which is <math>U⋅(M⋅U)^{-1}</math>:

Then find <math>M⁺</math> which is <math>U⋅(M⋅U)^{-1}</math>:

<math>

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</math>

Reading the columns from <math>T</math>, the first one confirms our period of 2/1, and the second column gives our generator 3/2. Which is unsurprising. In cents, that's 1200¢ × log₂(3/2) ≈ 701.955¢. The next unchanged interval will give a more interesting result.

Reading the columns from <math>M⁺</math>, the first one confirms our period of 2/1, and the second column gives our generator 3/2. Which is unsurprising. In cents, that's 1200¢ × log₂(3/2) ≈ 701.955¢. The next unchanged interval will give a more interesting result.

So let's do 5/4 now. We prepare a matrix out of these two unchanged intervals, 2/1 and 5/4, and call it <math>U</math>:

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</math>

Then find <math>T</math> which is <math>U⋅(M⋅U)^{-1}</math>:

Then find <math>M⁺</math> which is <math>U⋅(M⋅U)^{-1}</math>:

<math>

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</math>

Then find <math>T</math> which is <math>U⋅(M⋅U)^{-1}</math>:

Then find <math>M⁺</math> which is <math>U⋅(M⋅U)^{-1}</math>:

<math>

@@ Line 49: / Line 49: @@
 \end{bmatrix}
 </math>
+One property of an unchanged interval of a tuning is that it is [https://mathworld.wolfram.com/Eigenvector.html eigenvector] of the projection matrix<ref>For now, the best explanation of projection matrices seems to be on the [[fractional monzo]]s page.</ref> for the tuning where the eigenvalue <span><math>λ</math></span> is 1. If the projection matrix is <span><math>P</math></span>, by the definition of eigenvectors, that means <span><math>P⋅U = λ⋅U</math></span>, or <span><math>P⋅U = 1⋅U</math></span>, or simply <span><math>P⋅U = U</math></span>. In other words, the projection matrix maps the interval to itself; it is unchanged by the tuning. Because we know what <span><math>U</math></span> is, we could solve for <span><math>P</math></span> now. But we don't want <span><math>P</math></span>; we want the generators. Fortunately, <span><math>P</math></span> is defined in terms of our desired generators, <span><math>G</math></span>, and our mapping, <span><math>M</math></span>, like this: <span><math>P = GM</math></span>. Now, <span><math>G</math></span> is the [https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse pseudoinverse] of <span><math>M</math></span><ref>we can understand this fact in this context to mean that when you multiply <span><math>M</math></span> and <span><math>G</math></span> the other way around, <span><math>MG</math></span>, you get <span><math>I</math></span>, the identity matrix, and this is because <span><math>G</math></span> represents the generators in terms of the primes, just like any ordinary prime count vector ([[monzo]]), except often using fractional powers, so we map it with a mapping just like any ordinary interval, and if you look at the columns of <span><math>I</math></span> and think of them as generator count vectors, you can see that it represents our targets. In other words, <span><math>G</math></span> is the prime count vectors for which each one gets mapped to a different single one of the temperament's generators.</ref>; we often write the pseudoinverse of <span><math>M</math></span> as <span><math>M⁺</math></span>, so let's write <span><math>G</math></span> instead as <span><math>M⁺</math></span>. So if <span><math>P = M⁺⋅M</math></span>, then we can substitute that in for <span><math>P</math></span>, and our equation will now be <span><math>M⁺⋅M⋅U = U</math></span>. But we want to solve for our generators, so that's <span><math>M⁺</math></span>. So if we right multiply both sides by the inverse of <span><math>(M⋅U)</math></span>, we get <span><math>M⁺ = U(MU)^{-1}</math></span>; the <span><math>MU</math></span> on the left side cancels out. The rest is busywork.
 We multiply <span><math>M⋅U</math></span>:
@@ Line 77: / Line 79: @@
 </math>
-Then find <span><math>T</math></span> which is <span><math>U⋅(M⋅U)^{-1}</math></span>:
+Then find <span><math>M⁺</math></span> which is <span><math>U⋅(M⋅U)^{-1}</math></span>:
 <math>
@@ Line 96: / Line 98: @@
 </math>
-Reading the columns from <span><math>T</math></span>, the first one confirms our period of 2/1, and the second column gives our generator 3/2. Which is unsurprising. In cents, that's 1200¢ × log₂(3/2) ≈ 701.955¢. The next unchanged interval will give a more interesting result.
+Reading the columns from <span><math>M⁺</math></span>, the first one confirms our period of 2/1, and the second column gives our generator 3/2. Which is unsurprising. In cents, that's 1200¢ × log₂(3/2) ≈ 701.955¢. The next unchanged interval will give a more interesting result.
 So let's do 5/4 now. We prepare a matrix out of these two unchanged intervals, 2/1 and 5/4, and call it <span><math>U</math></span>:
@@ Line 135: / Line 137: @@
 </math>
-Then find <span><math>T</math></span> which is <span><math>U⋅(M⋅U)^{-1}</math></span>:
+Then find <span><math>M⁺</math></span> which is <span><math>U⋅(M⋅U)^{-1}</math></span>:
 <math>
@@ Line 193: / Line 195: @@
 </math>
-Then find <span><math>T</math></span> which is <span><math>U⋅(M⋅U)^{-1}</math></span>:
+Then find <span><math>M⁺</math></span> which is <span><math>U⋅(M⋅U)^{-1}</math></span>:
 <math>